What is the derivative of 2^{3x+1} ?

The derivative of 2^{3x+1} is 3ln(2)*2^{3x+1}

What is the first derivative of 2^{3x+1} ?

The first derivative of 2^{3x+1} is 3ln(2)*2^{3x+1}

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derivative of

2^{3 x + 1}

3 ln (2) \cdot 2^{3 x + 1}

Solution steps

\frac{d}{d x} (2^{3 x + 1})

Apply exponent rule:

a^{b} = e^{b l n (a)}

2^{3 x + 1} = e^{(3 x + 1) l n (2)}

= \frac{d}{d x} (e^{(3 x + 1) l n (2)})

Apply the chain rule:

e^{(3 x + 1) l n (2)} \frac{d}{d x} ((3 x + 1) ln (2))

= e^{(3 x + 1) l n (2)} \frac{d}{d x} ((3 x + 1) ln (2))

\frac{d}{d x} ((3 x + 1) ln (2)) = 3 ln (2)

= e^{(3 x + 1) l n (2)} \cdot 3 ln (2)

e^{(3 x + 1) l n (2)} = 2^{3 x + 1}

= 3 ln (2) \cdot 2^{3 x + 1}

\frac{d}{d r} (\frac{r}{r ^{2} + 1})

f (x) = x \cdot 5^{x}, at x = 1

\int 1 - x d x

y^{^{''}} - 3 y^{^{'}} + 2 y = 2 e^{3} t

x \to \infty lim (\frac{8}{x})

What is the derivative of 2^{3x+1} ?
The derivative of 2^{3x+1} is 3ln(2)*2^{3x+1}
What is the first derivative of 2^{3x+1} ?
The first derivative of 2^{3x+1} is 3ln(2)*2^{3x+1}